- #1
Apashanka
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if ##\bar x =\frac{\int x dv}{\int dv}=0##
Is necessarily ##\bar {x^2}=0##??
Is necessarily ##\bar {x^2}=0##??
What do you think and why?Apashanka said:Is the reverse true, e.g if ##\bar {x^2}=0## then ##\bar x=0##??, given ##x>0## and
##\bar x=\frac{\int x dv}{\int dv}##
Because I came across a situation where it is given that as ##\bar {A^2}##=0 ,so ##\bar {\vec A}##=0,that why I am askingNugatory said:What do you think and why?
"Situation"? That could only be true if all values of A are zero or complex, I think.Apashanka said:Because I came across a situation where it is given that as ##\bar {A^2}##=0 ,so ##\bar {\vec A}##=0,that why I am asking
\bar{}
is terrible in ##\LaTeX##. Consider using \overline{}
instead.The equation ##\overline{A_i^2}=0## means thatApashanka said:For ##\vec A## if ##\overline {A^2}=0## implies that ##\overline{A_i^2}=0## using this ##\frac{A_i \int_V A_i dv}{\int_V dv}-\frac{\int_V (\frac{dA_i}{dv}\int_V(A_i dv))dv}{\int_V dv}(=\bar A_i \int_V dA_i)=0## can this be used to proof that ##\bar A_i=0??##
(Average is taken over a volume V and ##\bar A_i=\frac{\int_V A_i dv}{\int_V dv}##)
What kind of Mathematical object is your ##A_i##?Apashanka said:For ##\vec A## if ##\overline {A^2}=0## implies that ##\overline{A_i^2}=0## using this ##\frac{A_i \int_V A_i dv}{\int_V dv}-\frac{\int_V (\frac{dA_i}{dv}\int_V(A_i dv))dv}{\int_V dv}(=\bar A_i \int_V dA_i)=0## can this be used to proof that ##\bar A_i=0??##
(Average is taken over a volume V and ##\bar A_i=\frac{\int_V A_i dv}{\int_V dv}##)
##A_i## are components of a vectorWWGD said:What kind of Mathematical object is your ##A_i##?
But how do you define the integral in this case? Is it given as a Real- or Otehrwise- valued function?Apashanka said:##A_i## are components of a vector
It's real and of course positive which are functions of ##f(x,y,z,t)##WWGD said:But how do you define the integral in this case? Is it given as a Real- or Otehrwise- valued function?
EDIT: Well, if the (Riemann) integral of a positive Real-valued function is 0, then the function must be identically 0. Otherwise, if ## f \geq \epsilon \geq 0 ## then ##\int_{[a,b]} f \geq \epsilon(b-a) >0 ##Apashanka said:It's real and of course positive which are functions of ##f(x,y,z,t)##
What I am just thinking( say in cartesian coordinates) is ##\int_v f(x,y,z)dv=\int_v f(x,y,z)dxdydz=\int_x f_x(x)dx\int_y f_y(y)\int_z f_z(z)dz##this expression is zero if each of the area under the curve on the ##x,y,z## axis is 0 ,or if it is having a point of zero crossing.WWGD said:EDIT: Well, if the (Riemann) integral of a positive Real-valued function is 0, then the function must be identically 0. Otherwise, if ## f \geq \epsilon \geq 0 ## then ##\int_{[a,b]} f \geq \epsilon(b-a) >0 ##
EDIT2: At most f can be non-zero at a set of measure 0.
Besides Dale's example: Wouldn't that imply that a mean of 0 would imply a variance of 0? That would be really odd, I'd say.Apashanka said:if ##\bar x =\frac{\int x dv}{\int dv}=0##
Is necessarily ##\bar {x^2}=0##??
Yes it's variance is then coming 0haushofer said:Besides Dale's example: Wouldn't that imply that a mean of 0 would imply a variance of 0? That would be really odd, I'd say.
I don't think you can always find a primitive. But if you did, since f (x,y,z) is Real valued, it would evaluate asApashanka said:What I am just thinking( say in cartesian coordinates) is ##\int_v f(x,y,z)dv=\int_v f(x,y,z)dxdydz=\int_x f_x(x)dx\int_y f_y(y)\int_z f_z(z)dz##this expression is zero if each of the area under the curve on the ##x,y,z## axis is 0 ,or if it is having a point of zero crossing.
For the function to be positive valued the first condition above is to be satisfied,isn't it??
Variance 0 implies all values are equal.Apashanka said:Yes it's variance is then coming 0
I don't think you can always find a primitive. But if you did, since f (x,y,z) is Real valued, it would evaluate asApashanka said:What I am just thinking( say in cartesian coordinates) is ##\int_v f(x,y,z)dv=\int_v f(x,y,z)dxdydz=\int_x f_x(x)dx\int_y f_y(y)\int_z f_z(z)dz##this expression is zero if each of the area under the curve on the ##x,y,z## axis is 0 ,or if it is having a point of zero crossing.
For the function to be positive valued the first condition above is to be satisfied,isn't it??
Expanding on this, yes, if your volume is zero, you can see it as having "degenerate" volume, i.e., a figure that is not strictly 3D ( under the same assumption that f(x,y,z) is non-negative), and that allows for degeneracy along different dimensions.WWGD said:I don't think you can always find a primitive. But if you did, since f (x,y,z) is Real valued, it would evaluate as
F(x_1,y_1,z_1)-F(x_0,y_0,z_0)
Variance 0 implies all values are equal.
Not necessarily, think about sin(x) and cos(x)Apashanka said:if ##\bar x =\frac{\int x dv}{\int dv}=0##
Is necessarily ##\bar {x^2}=0##??
Yes, that is what we did. Take X~N(0,1). Then ##E[X]=0, E[X^2]=E[(X-0)^2]=1 \neq E[X]^2##epenguin said:For the initial question isn't this one of those cases where tn suffices to think of a simple example? For example for the mean of two quantities to be zero they have to be equal and opposite sign. The squares are also equal but they are not opposite. Proved except for unineresting qualification where everything is zero.
The mean of x being 0 means that when all the values of x are added together and divided by the number of values, the result is 0. In other words, the average value of x is 0.
No, the mean of x being 0 does not guarantee that the mean of x-squared will also be 0. This is because the mean of x-squared is calculated by squaring each value of x and then finding the mean, which can result in a non-zero value even if the mean of x is 0.
The mean of x-squared is related to the mean of x through the variance of x. The variance of x is equal to the mean of x-squared minus the square of the mean of x. So if the mean of x is 0, the variance of x will also be 0, but the mean of x-squared may still be non-zero.
Yes, it is possible for the mean of x and the mean of x-squared to be equal even if the mean of x is not 0. This can occur if the values of x are symmetrically distributed around a non-zero mean.
If the mean of x is 0, the distribution of x-squared will have a smaller spread compared to the distribution of x. This is because squaring the values of x will result in all positive values, which will reduce the variability of the data. However, the shape of the distribution may still be affected by other factors such as the sample size and the underlying distribution of x.